An object with a mass of $16 k g$ $16 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{7}{8} m$ $7/8 m$ . If the spring's constant is $12 \frac{k g}{s^{2}}$ $12 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2016-04-20T08:01:30

$0.067$ $0.067$

The force exerted by a spring with spring constant $k$ $k$ and after a compression of $x$ $x$ is given as $- k x$ $-kx$ .

Now, as friction is always in the opposite direction to the applied force, hence, we have

$μ N = k x$ $muN = kx$

where $N$ $N$ is the normal force $= m g$ $= mg$

hence,

$μ = \frac{k x}{m g} = \frac{12 \cdot \frac{7}{8}}{16 \cdot 9.8} \approx 0.067$ $mu = (kx)/(mg) = (12*7/8)/(16*9.8) ~~0.067$

An object with a mass of 16kg16 kg is lying still on a surface and is compressing a horizontal spring by 78m7/8 m. If the spring's constant is 12kgs212 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?